# physics

A student throws a ball vertically upward such that it travels 7.4m to its maximum height.

Part A
If the ball is caught at the initial height 2.7s after being thrown, what is the ball’s average speed?

Part B
What is the ball’s average velocity?

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1. Part A
d=r*t
distance=rate*time

7.4m=r*2.7s

7.4m/2.7s=r

r=2.7m/s

Part B
Use the following equation:

Where:

Vf=0
a=g=9.8m/s
Vi=?
d=7.4m

0=Vi^2 +2(-9.8m/s^2)*(7.4m)

0=Vi^2-145m^2/s^2

145m^2/s^2=Vi^2

sqrt*(145)=Vi

Vi=12.0 m/s

AverageV=(Vf+Vi)/2=(0+12.0m/s)/2=6m/s

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2. (B) since the ball ended up where it started, its average velocity was 0

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3. Steve you may be correct, but both questions can be worded a little bit better. For A: I wasn't sure if d=7.4m or d=2(7.4)=14.8m. If so, then r=14.8m/2.7s=5.5m/s. For B: I wasn't sure if they wanted to know the average velocity for the trip up, or the trip up and back down. If so, then the average velocity is 0.

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4. Yeah. I've noticed that either textbooks have gotten very sloppy in their language, or students who post here aren't very careful in transcribing their assignments.

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5. Yes, but the for A the speed is 5.5m/s and 0 for the velocity. I must have been tired. I just didn' t read the question carefully.

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